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3 years ago

Please help me answer the question in the picture!

Mathematics

1 answer:

Slav-nsk [51]3 years ago

4 0

Answer:

x = 5 units.

Step-by-step explanation:

In a parallelogram. diagonals bisect each other.

This means $\overline{CE} = \overline{BE}$ .

$\implies 15 - x = 2x$

$\implies 15 = 3x$

$\implies x = 5$ .

Therefore, the length of x = 5 units.

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3 years ago

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The cost (c) to hire a dog trainer is varies directly with the amount of time (h) in hours spent training the dog.

NNADVOKAT [17]

Answer:

Part A) The proportional equation is $c=55h$

Part B) 8 hours of training

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $y/x=k$ or $y=kx$

Let

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h ----> the amount of time in hours

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therefore

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3 years ago

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4 years ago

Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5

alex41 [277]

The smallest such number is 1055.

We want to find $x$ such that

$\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}$

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

$x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}$

$x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}$

$x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}$

Taking everything together, we end up with the system

$\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}$

Now the moduli are coprime and we can apply the CRT.

We start with

$x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5$

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2$

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

$x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3$

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

$x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5$

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

$x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7$

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since $2\cdot4\equiv8\equiv1\pmod7$ , the inverse of 2 is 4.

So, we have to adjust $x$ to

$x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845$

and from the CRT we find

$x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}$

so that the general solution $x=210n+5$ for all integers $n$ .

We want a 4 digit solution, so we want

$210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5$

which gives $x=210\cdot5+5=1055$ .

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3 years ago